A Note The reader may note that in Example , we have used first derivative test instead of the second derivative test as the former is easy and short. Example Let AP and BQ be two vertical poles at points A and B, respectively. If AP = m, BQ = m and AB = m, then find the distance of a point R on AB from the point A such that RP + RQ is minimum. Solution Let R be a point on AB such that AR = x m.
Then RB = ( – x ) m (as AB = m). From Fig . , we have RP = AR + AP and RQ = RB + BQ Therefore RP + RQ = AR + AP + RB + BQ = x + ( ) + ( – x ) + ( ) = x – x + Let S ≡ S( x ) = RP + RQ = x – x + . Therefore S ′ ( x ) = x – .
Now S ′ ( x ) = gives x = . Also S ″ ( x ) = > , for all x and so S ″ ( ) > . Therefore, by second derivative test, x = is the point of local minima of S. Thus, the distance of R from A on AB is AR = x = m.
Example If length of three sides of a trapezium other than base are equal to 10cm, then find the area of the trapezium when it is maximum. Solution The required trapezium is as given in Fig . . Draw perpendiculars DP and Fig .
Fig . CQ on AB. Let AP = x cm. Note that ∆ APD ~ ∆ BQC.
Therefore, QB = x cm. Also, by Pythagoras theorem, DP = QC = . Let A be the area